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3 years ago

13

Tridil is infusing at 15 ml/hr on an infusion pump. The drug is mixed 50 mg in 500 ml D5W. How many MCG/minute is the patient re

ceiving?

1 answer:

olga nikolaevna [1]3 years ago

8 0

Answer:

patient receiving drug 25 MCG/minute

Explanation:

given data

infusing = 15 ml/hr

drug = 50 mg

D5W = 500 ml

to find out

How many MCG/minute

solution

we know infusing rate is 15 ml/hr = 0.25 ml/min

so 0.25 ml drug content = 50 /500 × 0.25

0.25 ml drug content = 0.025 mg

so here

rate of drug will be 0.025 mg

rate of drug = 0.025 mg = 25 × $10^{-6}$ gm/min

rate of drug = 25 MCG/minute

so patient receiving drug 25 MCG/minute

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Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

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1 year ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

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14.28 W

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Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

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3 years ago

Yuri [45]

Answer:

1) Adenine (A) 2. Uracil (U) 4) Cytosine (C)

Explanation:

"RNA consists of four nitrogenous bases: adenine, cytosine, uracil, and guanine."

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3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

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$\mu = \frac{m}{l}$

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The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

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$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

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The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

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$n=3$

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3 years ago

Margarita [4]

Answer: C

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In the absence of oxygen, combustion reaction can never take place. And when the supply of oxygen is being cut off, combustion reaction will automatically stop.

The correct option is C which is combustion

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4 years ago